jmc

We will consider the sum $A+B+C+D+E+F+G+H$. We know that the sum of any three consecutive terms is $30$ and that $C=5$, so $A+B+C=A+B+5=30$ and thus $A+B=25$. Now, we have $$A+B+C+D+E+F+G+H=A+(B+C+D)+(E+F+G)+H=A+30+30+H=A+H+60$$and $$A+B+C+D+E+F+G+H=(A+B)+(C+D+E)+(F+G+H)=25+30+30=85.$$Equating the two values we obtained for the sum, we find that $A+H+60=85$, so $A+H=\boxed{25}$.